-5t^2-4t+120=0

Solution for -5t^2-4t+120=0 equation:

-5t^2-4t+120=0

a = -5; b = -4; c = +120;
Δ = b2-4ac
Δ = -42-4·(-5)·120
Δ = 2416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{151}}{2*-5}=\frac{4-4\sqrt{151}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{151}}{2*-5}=\frac{4+4\sqrt{151}}{-10}
$